Which of the following real-valued functions | vedclass

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(C, D) function $f(x)$ is even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$.$(a)$ $f(x) = x^{3} \sin x$. Then $f(-x) = (-x)^{3} \sin(-x) = (-x^{3})(-\

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$f(x) = x - [x]$, where $[x]$ denotes the greatest integer less than or equal to $x$

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(C, D) function $f(x)$ is even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$.$(a)$ $f(x) = x^{3} \sin x$. Then $f(-x) = (-x)^{3} \sin(-x) = (-x^{3})(-\sin x) = x^{3} \sin x = f(x)$. Thus, $f(x)$ is even.$(b)$ $f(x) = x^{2} \cos x$. Then $f(-x) = (-x)^{2} \cos(-x) = x^{2} \cos x = f(x)$. Thus, $f(x)$ is even.$(c)$ $f(x) = e^{x} x^{3} \sin x$. Then $f(-x) = e^{-x} (-x)^{3} \sin(-x) = e^{-x} x^{3} \sin x 
eq f(x)$. Thus, $f(x)$ is not even.$(d)$ $f(x) = x - [x]$. Then $f(-x) = -x - [-x]$. Since $[-x] = -[x] - 1$ (for non-integer $x$), $f(-x) = -x - (-[x] - 1) = -x + [x] + 1 = 1 - (x - [x]) = 1 - f(x) 
eq f(x)$. Thus, $f(x)$ is not even.Therefore, options $(c)$ and $(d)$ are not even functions.

Which of the following real-valued functions is/are not even functions?

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